Produces cost equation

This relationship has been foimd to give reasonable results for individual pieces of equipment and for entire plants. Although, as shown by Williams (1947a,b), the exponent, m, may vary from 0.48 to 0.87 for equipment and from 0.38 to 0.90 for plants, the average value is close to 0.60. Accordingly, Eq. (16.3) is referred to as the six-tenths rule. Thus, if the capacity is doubled, the 0.6 exponent gives only a 52% increase in cost. Equation (16.3) is used in conjunction with Eq. (16.2) to take cost data from an earlier year at a certain capacity and estimate the current cost at a different capacity. As an example, suppose the total depreciable capital investment for a plant to produce 1,250 tonnes/day (1 tonne = 1,000 kg) of ammonia... 

A safety generalist became proficient in applied ergonomics. To convince management of the efficacy of his ergonomics proposals, his procedure for situations requiring expensive work modification was to produce cost-benefit data using only potential injury and illness cost savings in the benefit side of the equation. 

The amount of land required varies as well, not only as a function of the amount of production that is anticipated, but also on the type of culture system that is used. It may take several hectares of static culture ponds to produce the same biomass of animals as one modest size raceway through which large volumes of water are constantly flowed. Constmction costs vary from one location to another. Local labor and fuel costs must be factored into the equation. The experience of contractors in building aquaculture facihties is another factor to be considered. 

The second term in brackets in equation 36 is the separative work produced per unit time, called the separative capacity of the cascade. It is a function only of the rates and concentrations of the separation task being performed, and its value can be calculated quite easily from a value balance about the cascade. The separative capacity, sometimes called the separative power, is a defined mathematical quantity. Its usefulness arises from the fact that it is directly proportional to the total flow in the cascade and, therefore, directly proportional to the amount of equipment required for the cascade, the power requirement of the cascade, and the cost of the cascade. The separative capacity can be calculated using either molar flows and mol fractions or mass flows and weight fractions. The common unit for measuring separative work is the separative work unit (SWU) which is obtained when the flows are measured in kilograms of uranium and the concentrations in weight fractions. 

The annual processing cost Apo for a similar plant of a different size designed for an annual produc tion rate Ro can be approximately calculated from an equation of the form... 

Equation (9-214) gives the overpricing Acs per unit of produc tion for an increase in annual production rate AR. Equation (9-214) also gives the underpricing Acs per unit of production for a decrease in annual production rate AR. In the first case the fixed costs or overheads are said to be overabsorbed and in the second case underabsorbed. 

Later, when nitric acid was manufd from synthetic ammonia at relatively low cost, synthetic sodium nitrate was made from it either thru the reaction between nitric acid and soda ash, or by direct absorption of nitrogen dioxide in an aq soln of Na carbonate (see above equation). The Na nitrate-nitrite soln was then heated with excess nitric acid to convert the nitrite to nitrate, and the NO thus produced was recycled to the nitric acid plant (Ref 6)... 

Despite their flaws, batch processes have stood the test of time for a number of reasons, the most important of which is the flexibihty it brings to the manufacturer in terms of the range of products that the plant can produce, the feedstocks used to produce them, and the speed at which they can be brought to market with very limited information on physical properties, reaction kinetics, and so on (very few, if any, Michelin-starred chefs have ever measured the rheology or kinetics of their latest culinary creation). This flexibility, however, has a price which comes in the form of lower efficiencies in terms of production, energy, labor, and so on, and ultimately efficiency equates to cost However, one should never underestimate the pull of flexibility particularly, as discussed earlier in the examples of fermentation, where control of important parameters is difficult to achieve. 

In principle, the task of solving a linear algebraic systems seems trivial, as with Gauss elimination a solution method exists which allows one to solve a problem of dimension N (i.e. N equations with N unknowns) at a cost of O(N ) elementary operations [85]. Such solution methods which, apart from roundoff errors and machine accuracy, produce an exact solution of an equation system after a predetermined number of operations, are called direct solvers. However, for problems related to the solution of partial differential equations, direct solvers are usually very inefficient Methods such as Gauss elimination do not exploit a special feature of the coefficient matrices of the corresponding linear systems, namely that most of the entries are zero. Such sparse matrices are characteristic of problems originating from the discretization of partial or ordinary differential equations. As an example, consider the discretization of the one-dimensional Poisson equation... 

In this case, the wastewater is fed to the aerobic reactor where the remaining formaldehyde is oxidized to C02 (Equation 19.3) and urea is hydrolyzed to ammonia. This ammonia is then oxidized to nitrate (Equation 19.4). Nitrate goes to the denitrifying unit where it is reduced to dinitrogen gas in the presence of an electron donor, which is generally provided by organic matter (Equation 19.5). Because formaldehyde is oxidized in the first unit, methanol is commonly added to carry out this process, which produces an increase in operational costs. 

Ab initio calculations usually begin with a solution of the Hartree-Fock equations, which assumes the electronic wavefunction can be written as a single determinant of molecular orbitals. The orbitals are described in terms of a basis set of atomic functions and the reliability of the calculation depends on the quality of the basis set being used. Basis sets have been developed over the years to produce reliable results with a minimum of computational cost. For example, double zeta valence basis sets such as 3-21G [15] 4-31G [16] and 6-31G [17] describe each atom in the molecule with a single core Is function and two functions for the valence s and p functions. Such basis sets are commonly used, as there appears to be a cancellation of errors, which fortuitously allows them to predict quite accurate results. 

Sample Allocation for Estimating Concentration Levels. The variable cost of a sampling program that produces an estimate with variance given in equation (3) is... 

Solutions to Equations 7 produce solutions for K and J that are not whole numbers. To resolve this problem, we increase K to the next integer value and choose J so the cost constraint is exceeded by the smallest amount possible. 

Balanced chemical reactions are critical to the chemical industry. Plant managers must know the amount of reactants necessary to yield a product to keep production lines moving. If reactants or products are not pure, this must be taken into account. Alternative reactions can be examined to find the most cost-effective process. For example, a fertilizer may be produced using one process with sulfuric acid as a reactant or another process that starts with ammonium carbonate. Working through the balanced equations for each process would help decide which process to adopt. 

The Problem A businessman wants to know when the sale of a particular item reaches a profit level of 2,000. The revenue equation is R = 200x- 0.4X2, and the cost to produce x items is determined with C = 4,000 + lOOx. How many items have to be produced and sold to net a profit of 2,000 ... 

Using a well known result, for a linear demand curve, marginal revenue is MR = (-a/ ) + (2/p)q. The profit maximizing output is that at which marginal revenue equals marginal cost, or 10. Equating MR to 10 and solving for q produces q = a/2 + 5p, so we require a confidence interval for this combination of the parameters. 

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